Forget it, Jake. It's floating point > 999999999999999000000 999999999999999000000 > 999999999999999000000 % 1000000 951424
-
-
Replying to @qntm
Why print "999999999999999000000" when the actual float is exactly 999999999999998951424? Why conceal those extra digits?
6 replies 0 retweets 2 likes -
Replying to @qntm
The actual float is NOT exactly that value. It's that value +/- 2^20.
1 reply 0 retweets 0 likes -
-
Replying to @qntm
No, it really isn't. :P Any value in that range has the same floating-point representation. It's not 1:1.
1 reply 0 retweets 0 likes -
Replying to @codahighland
This value behaves like 999999999999998951424 when I do arithmetic with it. The value is 999999999999998951424.
1 reply 0 retweets 0 likes -
Replying to @qntm
No it doesn't. It behaves like 9.99999999999999e+20. Try adding 1 to it.
2 replies 0 retweets 0 likes -
Replying to @codahighland
999999999999998951424 + 1 is 999999999999998951425. This value cannot be exactly represented so it takes the closest, 999999999999998951424
1 reply 0 retweets 0 likes -
Replying to @qntm
That looks awfully like the behavior of 999999999999998951424 +/- 2^20 ;)
1 reply 0 retweets 0 likes
(999999999999998951424 +/- 2^20) + 1 should yield 999999999999998951425 +/- 2^20
-
-
Replying to @qntm @codahighland
The "error bar" you're describing is only in the result of the previous calculation (if any) which led to the current float
1 reply 0 retweets 0 likes - 2 more replies
New conversation -
Loading seems to be taking a while.
Twitter may be over capacity or experiencing a momentary hiccup. Try again or visit Twitter Status for more information.